Question

Resolve the given fraction into partial fractions
$\dfrac{{{\text{3}}{{\text{x}}^3} - 8{{\text{x}}^2} + 10}}{{{{\left( {{\text{x - 1}}} \right)}^4}}}$

Answer 1

Hint: To solve the given equation we apply the partial fraction decomposition method, i.e. we write out the partial fraction for each factor, then solve for coefficients by substituting zeroes in the equation.

Complete step-by-step answer:
Given Data,
$\dfrac{{{\text{3}}{{\text{x}}^3} - 8{{\text{x}}^2} + 10}}{{{{\left( {{\text{x - 1}}} \right)}^4}}}$ Can be resolved into partial fraction as $\dfrac{{\text{A}}}{{{\text{x - 1}}}} + \dfrac{{\text{B}}}{{{{\left( {{\text{x}} - 1} \right)}^2}}} + \dfrac{{\text{C}}}{{{{\left( {{\text{x - 1}}} \right)}^3}}} + \dfrac{{\text{D}}}{{{{\left( {{\text{x - 1}}} \right)}^4}}}$
So, ${\text{3}}{{\text{x}}^3} - 8{{\text{x}}^2} + 10$= ${\text{A}}{\left( {{\text{x - 1}}} \right)^3} + {\text{B}}{\left( {{\text{x - 1}}} \right)^2} + {\text{C}}\left( {{\text{x - 1}}} \right) + {\text{D}}$
Put x = 1 in the above equation we get 3 – 8 + 10 = D
⟹D = 5.
${\text{3}}{{\text{x}}^3} - 8{{\text{x}}^2} + 10$= ${\text{A}}\left( {{{\text{x}}^3} - 1 - 3{{\text{x}}^2}{\text{ + 3x}}} \right) + {\text{B}}\left( {{{\text{x}}^2}{\text{ + 1 - 2x}}} \right) + {\text{C}}\left( {{\text{x - 1}}} \right) + {\text{D}}$
$\left( {{{\left( {{\text{a - b}}} \right)}^3} = {{\text{a}}^3} - {{\text{b}}^3} - {\text{3}}{{\text{a}}^2}{\text{b + 3a}}{{\text{b}}^2}{\text{ and }}{{\left( {{\text{a - b}}} \right)}^2} = {{\text{a}}^2} + {{\text{b}}^2} - 2{\text{ab}}} \right)$
Now comparing the coefficients of ${{\text{a}}^3}$on both sides, we get, A = 3.
So the equation reduces to ${\text{3}}{{\text{x}}^3} - 8{{\text{x}}^2} + 10$= $3\left( {{{\text{x}}^3} - 1 - 3{{\text{x}}^2}{\text{ + 3x}}} \right) + {\text{B}}\left( {{{\text{x}}^2}{\text{ + 1 - 2x}}} \right) + {\text{C}}\left( {{\text{x - 1}}} \right) + {\text{D}}$
Now compare the constant on both sides we get, 10 = -3 + B – C + 5
⟹8 = B – C --- (1)
Comparing the coefficient of ${{\text{x}}^2}$, we get -8 = -9 + B
⟹B = 1
Put B = 1 in (1) we get C = -7.
∴ $\dfrac{{{\text{3}}{{\text{x}}^3} - 8{{\text{x}}^2} + 10}}{{{{\left( {{\text{x - 1}}} \right)}^4}}}$= $\dfrac{3}{{{\text{x - 1}}}} + \dfrac{1}{{{{\left( {{\text{x}} - 1} \right)}^2}}} - \dfrac{7}{{{{\left( {{\text{x - 1}}} \right)}^3}}} + \dfrac{5}{{{{\left( {{\text{x - 1}}} \right)}^4}}}$

Note: In order to solve this type of question it is essential to know the steps involved in the partial fraction decomposition method. In algebra, the partial fraction decomposition or partial fraction expansion of a rational function is an operation that consists of expressing the fraction as a sum of a polynomial and one or several fractions with a simpler denominator.